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Chapter 8: Problem 85

Jeff of the Jungle swings on a \(7.6-\mathrm{m}\) vine that initially makes an angle of \(37^{\circ}\) with the vertical. If Jeff starts at rest and has a mass of \(78 \mathrm{kg}\), what is the tension in the vine at the lowest point of the swing?

Short Answer

Expert verified
The tension in the vine at the lowest point is approximately 1076.1 N.

Step by step solution

01

Identify the Forces at the Lowest Point

At the lowest point of the swing, the two main forces acting on Jeff are the gravitational force (his weight, downward) and the tension in the vine (upward). The acceleration due to gravity is given as \(g = 9.8 \, \text{m/s}^2\). The gravitational force is: \[ F_\text{gravity} = mg = 78 \, \text{kg} \times 9.8 \, \text{m/s}^2.\]
02

Calculate Gravitational Force

Substitute the known values into the formula for gravitational force to calculate it. \[ F_\text{gravity} = 78 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 764.4 \, \text{N}.\] So, the gravitational force acting on Jeff is 764.4 N.
03

Determine Potential Energy at the Start

Initially, Jeff is at rest at a height above the lowest point. Calculate the initial potential energy at the start. The increase in height \(h\) can be found using the formula: \[ h = L - L\cos(\theta) = 7.6 \, \text{m} - 7.6 \, \text{m} \times \cos(37^\circ).\]
04

Calculate the Height Change

Calculate the numerical value of height change using the cosine of 37 degrees: \[ h = 7.6 \, \text{m} - 7.6 \, \text{m} \times 0.7986 \approx 1.53 \, \text{m}.\] This is the height Jeff falls through.
05

Calculate Initial Potential Energy

The initial potential energy can be calculated with: \[ PE_\text{initial} = mgh = 78 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 1.53 \, \text{m} \approx 1173.8 \, \text{J}.\]
06

Convert Potential Energy to Kinetic Energy

At the lowest point, the potential energy becomes kinetic energy. Therefore, \[ KE = PE_\text{initial} = 1173.8 \, \text{J}.\] The kinetic energy at the bottom is used to find Jeff's speed.
07

Calculate Speed at the Lowest Point

Use the kinetic energy formula to find velocity: \[ KE = \frac{1}{2}mv^2 \Rightarrow 1173.8 \, \text{J} = \frac{1}{2} \times 78 \, \text{kg} \times v^2.\] Solve for \(v\): \[ v^2 = \frac{2 \times 1173.8}{78} \Rightarrow v \approx 5.48 \, \text{m/s}.\]
08

Calculate Centripetal Force

At the lowest point, centripetal force is required for motion around the circular path: \[ F_\text{centripetal} = \frac{mv^2}{L} = \frac{78 \, \text{kg} \times (5.48 \, \text{m/s})^2}{7.6 \, \text{m}} \approx 311.7 \, \text{N}.\]
09

Combine Forces to Find Tension

The tension at the lowest point must counteract both gravity and provide centripetal force: \[ T = F_\text{gravity} + F_\text{centripetal} = 764.4 \, \text{N} + 311.7 \, \text{N} = 1076.1 \, \text{N}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Force
Understanding gravitational force is crucial when analyzing a pendulum like Jeff's vine. Gravitational force is the pull that the Earth exerts on an object, which in Jeff's case, is directed downwards toward the center of the Earth.
The formula to calculate gravitational force is: \[ F_{\text{gravity}} = mg \]where:
  • \( m \) is the mass of the object,
  • \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \).

In Jeff's problem, his mass is 78 kg, so multiplying by the gravitational constant gives a gravitational force of 764.4 N pulling him towards the Earth. This force is ever-present and plays a pivotal role in how much tension is required in the vine, as it affects the motion dynamics of the swing.
Centripetal Force
Centripetal force is the necessary force that keeps an object moving in a circular path. For Jeff's swing, this means the force needed to keep him moving in the arc of his swing.
At the lowest point of the swing, this force is directed towards the center of the swing's path, opposite to the direction of gravitational force.
The formula for centripetal force is: \[ F_{\text{centripetal}} = \frac{mv^2}{L} \]where:
  • \( m \) is the mass of the object,
  • \( v \) is the velocity at the lowest point,
  • \( L \) is the length of the swing.

Given Jeff's speed at the lowest point is 5.48 m/s and the vine is 7.6 m long, the centripetal force needed is approximately 311.7 N. This force, alongside gravitational force, affects the tension required in the vine for a safe and complete swing.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. In a pendulum, kinetic energy is at its maximum at the lowest point in the swing, where potential energy converts entirely to kinetic energy.
The formula for kinetic energy is: \[ KE = \frac{1}{2}mv^2 \]where:
  • \( m \) is the mass of the object,
  • \( v \) is the velocity at that point.

In Jeff's case, once he drops to the lowest point, all the potential energy he initially had at the top is now kinetic energy. This is calculating to be about 1173.8 J. Understanding kinetic energy helps in determining how fast Jeff is moving at the lowest point, which is a critical factor in calculating the tension in the vine necessary to counteract his motion and gravitational pull.
Potential Energy
Potential energy in a pendulum is primarily gravitational potential energy, related to the height of the pendulum above its lowest point.
The formula to calculate it is:\[ PE = mgh \]where:
  • \( m \) is the mass,
  • \( g \) is the acceleration due to gravity,
  • \( h \) is the height above the reference point.

Initially, Jeff starts at an angle, which means he is elevated compared to the lowest point, giving him potential energy. As he swings downward, this energy is converted into kinetic energy. The initial potential energy of Jeff is about 1173.8 J, calculated from a height difference of approximately 1.53 m. This value gives insight into the dynamics of energy conversion during Jeff's swinging motion.

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Most popular questions from this chapter

A block of mass \(m=0.95 \mathrm{kg}\) is connected to a spring of force constant \(k=775 \mathrm{N} / \mathrm{m}\) on a smooth, horizontal surface. (a) Plot the potential energy of the spring from \(x=-5.00 \mathrm{cm}\) to \(x=5.00 \mathrm{cm}\) (b) Determine the turning points of the block if its speed at \(x=0\) is \(1.3 \mathrm{m} / \mathrm{s}\)

An \(81.0-k g\) in-line skater does +3420 J of nonconservative work by pushing against the ground with his skates. In addition, friction does -715 J of nonconservative work on the skater. The skater's initial and final speeds are \(2.50 \mathrm{m} / \mathrm{s}\) and \(1.22 \mathrm{m} / \mathrm{s},\) respectively. (a) Has the skater gone uphill, downhill, or remained at the same level? Explain. (b) Calculate the change in height of the skater.

A \(1.9-\mathrm{kg}\) block slides down a frictionless ramp, as shown in Figure \(8-28 .\) The top of the ramp is \(1.5 \mathrm{m}\) above the ground; the bottom of the ramp is \(0.25 \mathrm{m}\) above the ground. The block leaves the ramp moving horizontally, and lands a horizontal distance \(d\) away. Find the distance \(d\).

\(A\) 15,800-kg truck is moving at \(12.0 \mathrm{m} / \mathrm{s}\) when it starts down a \(6.00^{\circ}\) incline in the Canadian Rockies. At the start of the descent the driver notices that the altitude is \(1630 \mathrm{m}\). When she reaches an altitude of \(1440 \mathrm{m},\) her speed is \(29.0 \mathrm{m} / \mathrm{s}\). Find the change in (a) the gravitational potential energy of the system and (b) the truck's kinetic energy. (c) Is the total mechanical energy of the system conserved? Explain.

A 2.9-kg block slides with a speed of 1.6 \(\mathrm{m} / \mathrm{s}\) on a frictionless horizontal surface until it encounters a spring. (a) If the block compresses the spring \(4.8 \mathrm{cm}\) before coming to rest, what is the force constant of the spring? (b) What initial speed should the block have to compress the spring by \(1.2 \mathrm{cm} ?\)
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